Suspension of a circle is a sphere Define the suspension of a topological space as 
$$SX=S\times I/\sim$$
where $\sim$ is the relation that identifies points of the form $(x,0)$ with one point and the ones of the form $(x,1)$ with another. When taking $X=S^1$, $SS^1$ looks like two cones glued by the unit cicle on the $XY$ plane (the Wikipedia article has a more illustrative pictur). The resulting space looks already homeomorphic to the sphere and I thought there might be a nice proof involving just the planar representation of compact surfaces, in this case the cylinder, so here is my try:

Identify $S^1\times I$ as the unit square with the vertical sides identified. Now the equivalence relation results on colapsing the boundary of the cylinder (to copies of $S^1$). In the planar model that should mean that the horizontal sides collapse into one point each, so that is just the unit disk $D^2$ with its boundary $\partial D^2=S^1$ identified with a single point. Since $D^2/S^1 \cong S^2$, then $SS^1\cong S^2$.

Does this work, or am I missing something? Also, is there another proof of this result? And is this generalizable to more dimensions? Thank you in advance.
 A: There is a bit of a jump from saying (a) you take $S^1\times I$ and collapse both ends each to a single point to (b) this space is $D^2$ with its boundary collapsed to a single point.  You are describing a reduced suspension of $S^1$ that way.
To show that a space is $S^2$, demonstrate a homeomorphism.  The more hand-wavey "this is this" sort of reasoning is for when you know you could write down an explicit homeomorphism when pressed.
Consider a map $f:S^1\times I\to S^2$ defined by $f((x,y),t)=(x\sqrt{1-t^2},y\sqrt{1-t^2},t)$, with $S^1$ as the unit circle in $\mathbb{R}^2$, with $I=[-1,1]$, and with $S^2$ as the unit sphere in $\mathbb{R}^3$.  This is continuous with continuous inverse for points in $S^1\times(-1,1)$ since one could write down the inverse, and since the function is made of continuous functions on this domain.  For an open neighborhood $U$ of $(0,0,1)$ in $S^2$, replace $U$ with an open ball centered at $(0,0,1)$ that is contained within $U$.  One can check $f^{-1}(U)$ is $S^1\times(1-\varepsilon,1]$ for some $\varepsilon$, so it follows that $f$ is continuous on $S^1\times \{1\}$.  Similar reasoning applies for $(0,0,-1)$.  Therefore, $f$ is indeed continuous.
Additionally, $f$ is a quotient map, which we basically already checked.
The fibers of $f$ are $S^1\times\{-1\}$, $S^1\times\{1\}$, and $\{p\}\times \{t\}$ for all $p\in S^1$ and $-1<t<1$.  This is the equivalence relation for a suspension, so therefore $S^2$ is the suspension of $S^1$.
This argument generalizes easily to $S(S^n)\approx S^{n+1}$.

With more experience with CW complexes and the classification of surfaces, you can make your square argument more precise.  Give $S^1$ the CW structure of one $0$-cell and one $1$-cell, and give $I$ the CW structure of two $0$-cells and one $1$-cell.  $S^1\times I$ has a natural CW structure of two $0$-cells, three $1$-cells, and one $2$-cell.  Collapsing  the two ends of this cylinder is a well-defined operation on CW complexes, yielding a space with two $0$-cells, one $1$-cell, and one $2$-cell.  At this point, one can recognize this is a closed surface of genus zero, so it is a sphere.  Or, recognize there is a disk neighborhood around the $1$-cell, collapse it, and you have the standard complex for $S^2$, which is a $2$-cell whose boundary is attached to a $0$-cell.
